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Last updated on September 15, 2025
The derivative of e^(x²) is used to determine how the function e^(x²) changes in response to a slight change in x. Derivatives are crucial in calculating various real-life situations, like rates of change in physics or cost functions in economics. We will now discuss the derivative of e^(x²) in detail.
We aim to understand the derivative of e^(x²). It is commonly represented as d/dx (e^(x²)) or (e^(x²))', and its value is 2x·e^(x²).
The function e^(x²) has a clear derivative, indicating that it is differentiable over its entire domain. Key concepts include:
Exponential Function: e^(x²), where e is the base of natural logarithms.
Chain Rule: A rule for differentiating composite functions, essential for e^(x²).
Product Rule: Used in some methods for differentiation, though not directly for e^(x²).
The derivative of e^(x²) is denoted as d/dx (e^(x²)) or (e^(x²))'.
The formula used to differentiate e^(x²) is: d/dx (e^(x²)) = 2x·e^(x²)
This formula applies to all x in the real number domain, as e^(x²) is defined everywhere.
The derivative of e^(x²) can be derived using different methods. Here, we'll use the chain rule and the first principle to demonstrate it:
To prove the differentiation of e^(x²) using the chain rule: Let u = x², then the function becomes e^u. The derivative of e^u with respect to u is e^u. The derivative of u = x² with respect to x is 2x.
By the chain rule, d/dx (e^(x²)) = e^u · du/dx = e^(x²) · 2x = 2x·e^(x²).
We can also derive the derivative using the first principle, which expresses the derivative as the limit of the difference quotient. Let f(x) = e^(x²).
Its derivative can be expressed as: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [e^((x+h)²) - e^(x²)] / h = limₕ→₀ [e^(x² + 2xh + h²) - e^(x²)] / h = limₕ→₀ [e^(x²) · (e^(2xh + h²) - 1)] / h
Using the expansion of e^u for small u, e^u ≈ 1 + u, we have: = limₕ→₀ [e^(x²) · (2xh + h²)] / h = limₕ→₀ [e^(x²) · (2x + h)] = e^(x²) · 2x
Hence, the derivative of e^(x²) is 2x·e^(x²).
When a function is differentiated multiple times, the subsequent derivatives are called higher-order derivatives. Understanding these derivatives provides deeper insights into the function's behavior.
For the first derivative of a function, we denote it as f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x), which shows the rate of change of the rate of change. Similarly, the third derivative, f′′′(x), results from the second derivative, and this pattern continues.
For the nth derivative of e^(x²), we generally use fⁿ(x) to denote the nth derivative of a function f(x), which tells us about the change in the rate of change, continuing for higher-order derivatives.
Since e^(x²) is defined for all real x, there are no points where the derivative is undefined.
At x = 0, the derivative of e^(x²) = 2x·e^(x²) simplifies to 0·e^(0) = 0, indicating no change at that point.
Students frequently make mistakes when differentiating e^(x²). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of e^(x²)cos(x).
Here, we have f(x) = e^(x²)cos(x).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^(x²) and v = cos(x).
Let’s differentiate each term, u′ = d/dx (e^(x²)) = 2x·e^(x²) v′ = d/dx (cos(x)) = -sin(x)
Substituting into the given equation, f'(x) = (2x·e^(x²))(cos(x)) + (e^(x²))(-sin(x))
Let’s simplify terms to get the final answer, f'(x) = 2x·e^(x²)cos(x) - e^(x²)sin(x)
Thus, the derivative of the specified function is 2x·e^(x²)cos(x) - e^(x²)sin(x).
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
A company's revenue is modeled by the function R(x) = e^(x²)x. Calculate the rate of change of revenue when x = 1.
We have R(x) = e^(x²)x (revenue function)...(1)
Now, we will differentiate the equation (1)
Using the product rule, R'(x) = d/dx (e^(x²)x) = (d/dx (e^(x²)))x + e^(x²)(d/dx (x)) = (2x·e^(x²))x + e^(x²)(1) = 2x²·e^(x²) + e^(x²)
Given x = 1, substitute this into the derivative: R'(1) = 2(1)²·e^(1²) + e^(1²) = 2·e + e = 3e
Thus, the rate of change of revenue at x = 1 is 3e.
We find the rate of change of revenue by differentiating the given revenue function using the product rule. We then substitute x = 1 into the derivative to get the final result.
Derive the second derivative of the function y = e^(x²).
The first step is to find the first derivative, dy/dx = 2x·e^(x²)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2x·e^(x²)]
Here we use the product rule, d²y/dx² = 2(e^(x²) + x(2x·e^(x²))) = 2e^(x²) + 4x²·e^(x²)
Therefore, the second derivative of the function y = e^(x²) is 2e^(x²) + 4x²·e^(x²).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate the expression again to find the second derivative.
Prove: d/dx (e^(x²)tan(x)) = 2x·e^(x²)tan(x) + e^(x²)sec²(x).
Let’s start using the product rule: Consider y = e^(x²)tan(x)
To differentiate, we use the product rule: dy/dx = d/dx (e^(x²))tan(x) + e^(x²)d/dx (tan(x)) = (2x·e^(x²))tan(x) + e^(x²)sec²(x)
Hence, proved.
In this step-by-step process, we use the product rule to differentiate the equation. We replace each term with its derivative and simplify to derive the equation.
Solve: d/dx (e^(x²)/x).
To differentiate the function, we use the quotient rule: d/dx (e^(x²)/x) = (d/dx (e^(x²))x - e^(x²)d/dx (x)) / x²
We will substitute d/dx (e^(x²)) = 2x·e^(x²) and d/dx (x) = 1: = (2x·e^(x²)x - e^(x²)·1) / x² = (2x²·e^(x²) - e^(x²)) / x² = 2x·e^(x²) - e^(x²)/x²
Therefore, d/dx (e^(x²)/x) = (2x²·e^(x²) - e^(x²)) / x².
In this process, we differentiate the given function using the quotient rule. We then simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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